[seqfan] Re: Exchange rightmost two base-4 digits of n>1; a(0)=0, a(1)=4. (base 4 analogue of what A181099 is in base 3, and A080412 is base 2
allouche at math.jussieu.fr
allouche at math.jussieu.fr
Tue Oct 12 19:02:10 CEST 2010
Yes indeed the first differences are periodic
(4, 4, 4, -11, 4, 4, 4, -11, 4, 4, 4, -11, 4, 4, 4, 1]
(not quite what you wrote).
The proof is easy: translating the very definition of the sequence
(let us call it u(n)) gives:
u(16n) = 16n
u(16n+1) = 16n+4
u(16n+2) = 16n+8
etc. (please pardon my lazyness in failing to write the next 13 equalities).
The claim about the periodicity of the first differences follows immediately.
best
jp
Georgi Guninski <guninski at guninski.com> a écrit :
[Cacher les citations]
On Mon, Oct 11, 2010 at 12:02:31AM -0700, Jonathan Post wrote:
Exchange rightmost two base-4 digits of n>1; a(0)=0, a(1)=4.
0, 4, 8, 12, 1, 5, 9, 13, 2, 6, 10, 14, 3, 7, 11, 15, 16, 20, 24, 28,
17, 21, 25, 29, 18, 22, 26, 30, 19, 23, 27, 31, 32, 36, 40, 44, 33,
37, 41, 45, 34, 38, 42, 46, 35, 39, 43, 47, 48, 52, 56, 60, 49, 53,
57, 61, 50
first differences seem periodic:
[4, 4, 4, -11, 4, 4, 4, -11, 4, 4, 4, 1, 4, 4, 4, -11, ...]
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